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  • One diagonal of a square is the intercept of the line <img alt="\mathrm{\frac{x}{a}+\frac{y}{b}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%7D%7Ba%7D&plus;%

One diagonal of a square is the intercept of the line \mathrm{\frac{x}{a}+\frac{y}{b}=1} between the axes. Find the coordinates of other two vertices.

 

Option: 1

\mathrm{ \left(\frac{a+b}{2}, \frac{a+b}{2}\right) \text { and }\left(\frac{b-a}{2}, \frac{a-b}{2}\right)}


Option: 2

\mathrm{\left(\frac{a+b}{2}, \frac{a+b}{2}\right) \text { and }\left(\frac{a-b}{2}, \frac{b-a}{2}\right) }


Option: 3

\mathrm{\left(\frac{a-b}{2}, \frac{a-b}{2}\right) \text { and }\left(\frac{a+b}{2}, \frac{a+b}{2}\right) }


Option: 4

\mathrm{\left(\frac{b-a}{2}, \frac{b-a}{2}\right) \text { and }\left(\frac{a+b}{2}, \frac{a+b}{2}\right)}


Answers (1)

best_answer

The coordinates of centre E are \mathrm{ \left(\frac{a}{2}, \frac{b}{2}\right)}
Slope of AC=\mathrm{ -\frac{b}{a}}
Slope of DB=\mathrm{ \frac{a}{b} \Rightarrow \tan \theta=\frac{a}{b} \Rightarrow \sin \theta=\frac{a}{\sqrt{a^2+b^2}}, \cos \theta=\frac{b}{\sqrt{a^2+b^2}} }Length \mathrm{A E=E C=E B=E D=\frac{I}{2} A C=\frac{l}{2} \sqrt{a^2+b^2}}
Using distance form for DB

\mathrm{\frac{x-\frac{a}{2}}{\frac{b}{\sqrt{a^2+b^2}}}=\frac{y-\frac{b}{2}}{\frac{a}{\sqrt{a^2+b^2}}}= \pm \frac{l}{2} \sqrt{a^2+b^2} }

\mathrm{\Rightarrow \quad x=\frac{a \pm b}{2}, y=\frac{b \pm a}{2} }
The coordinates of other two vertices are thus 

\mathrm{B\left(\frac{a+b}{2}, \frac{b+a}{2}\right) and~ D \left(\frac{a-b}{2}, \frac{b-a}{2}\right)}

Posted by

avinash.dongre

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